我有一个调查结果的R数据框。每列都是对调查问题的答复。它可以取值1到10和NA。我想把它变成频率表。
这是我拥有的数据的示例。我假装值从1到3,而不是1到10。
data.frame(
"Person" = c(1,2,3),
"Question1" = c(NA, "1", "1"),
"Question2" = c("1", "2", "3")
)
我想要什么:
data.frame(
"Question" = c("Question1", "Question2"),
"Frequency of 1" = c(2, 1),
"Frequency of 2" = c(0 , 1),
"Frequency of 3" = c(0, 1)
)
我已经尝试过使用likert包中的likert(),但得到的分数结果不正确。有解决这个问题的简单方法吗?
答案 0 :(得分:3)
这是使用dplyr和purrr软件包的解决方案
library(dplyr)
library(purrr)
data.frame(
"Person" = c(1,2,3),
"Question1" = c(NA, "1", "1"),
"Question2" = c("1", "2", "3")
)
df %>%
select(-Person) %>%
mutate_all(~ factor(.x, levels = as.character(1:10) ) %>% addNA() ) %>%
map(table) %>%
transpose() %>%
map(as.integer) %>%
set_names( ~ paste0("Frequency of ",ifelse(is.na(.), "NA", .))) %>%
as_tibble() %>%
mutate(Question = setdiff(names(df),"Person")) %>%
select(Question,everything(), "Frequency of NA" = `Frequency of ` )
答案 1 :(得分:2)
一种data.table
解决方案:
require(data.table)
setDT(df)
# Melt data:
df <- melt(df, id.vars = "Person", value.name = "Question")
# Cast data to required structure:
df <- data.frame(dcast(df, variable ~ Question))
# Rename variables and remove NA count (as per Ops question):
names(df)[1] <- "Question"
names(df)[-1] <- gsub("X", "Frequency of ", names(df)[-1])
df$NA. <- NULL
df
# Question Frequency of 1 Frequency of 2 Frequency of 3
#1 Question1 2 0 0
#2 Question2 1 1 1
或单行答案:
dcast(melt(setDT(df), id.vars="Person", value.name="Question")[!Question %in% NA][, Question := paste0("Frequency of ", Question)], variable ~ Question)
答案 2 :(得分:1)
另一种tidyverse
可能性是:
df %>%
gather(Question, val, -Person, na.rm = TRUE) %>%
group_by(Question, val) %>%
summarise(res = length(val)) %>%
ungroup() %>%
mutate(val = paste0("Frequency.of.", val)) %>%
spread(val, res, fill = NA)
Question Frequency.of.1 Frequency.of.2 Frequency.of.3
<chr> <int> <int> <int>
1 Question1 2 NA NA
2 Question2 1 1 1
首先,将数据从宽格式转换为长格式。其次,它根据问题计算频率。最后,它创建“ Frequency.of”。变量并将数据返回其所需的形状。
或者,如果您还想计算每个问题的NA值:
df %>%
gather(Question, val, -Person) %>%
group_by(Question, val) %>%
summarise(res = length(val)) %>%
ungroup() %>%
mutate(val = paste0("Frequency.of.", val)) %>%
spread(val, res, fill = NA)
Question Frequency.of.1 Frequency.of.2 Frequency.of.3 Frequency.of.NA
<chr> <int> <int> <int> <int>
1 Question1 2 NA NA 1
2 Question2 1 1 1 NA
答案 3 :(得分:0)
这不是最优雅的方法,但可能会有所帮助:df2是您的数据集。 数据:
df2<-data.frame(
"Person" = c(1,2,3),
"Question1" = c(NA, "1", "1"),
"Question2" = c("1", "2", "3"),stringsAsFactors = F
)
目标: 编辑::您可以按照以下步骤“自动化”
df2[is.na(df2)]<-0 #To allow numeric manipulation
values<-c("1","2","3")
Final_df<-sapply(values,function(val) apply(df2[,-1],2,function(x) sum(x==val)))
Final_df<-as.data.frame(Final_df)
names(Final_df)<-paste0("Frequency of_",1:ncol(Final_df))
这将产生:
Frequency of_1 Frequency of_2 Frequency of_3
Question1 2 0 0
Question2 1 1 1